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2 years ago

45 marbles in 9 bags; 150 marbles in 36 bags is this rate proportional?

Mathematics

1 answer:

stellarik [79]2 years ago

7 0

Answer:

No, the rate is not proportional.

Step-by-step explanation:

If there are 45 marbles in 9 bags, we can assume that there are 5 marbles in each bag.

Going by that rate, we can do 5 * 36 = 180, NOT 150.

Thus, the rate is not proportional.

Another way to solve would be to do 45/9 and 150/36. If the rate is proportional, the answer for both should be the same.

However, 45/9=5 and 150/36=4.16666666667.

Those two answers are NOT equal.

Thus, the rate is not proportional.

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There are 30 singers in a show. The ratio of singers to dancers is 5:6. How many dancers are there?

Tamiku [17]

Answer: 32 dancers

Step-by-step explanation:

I drew a chart and on the side with the singers you count by 5's until u get to 30. On the dancers side of the chart you count by 6's until it lines up with 30 on the singers side.

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Which function below is the inverse of f(x)= 3x+5/7

Ipatiy [6.2K]

Inverse
replace f(x) wih y
solve for x
replace x with f⁻¹(x) and y with x

so

f(x)=3x+5/7
replace
y=3x+5/7
solve for x
minnus 5/7 both sides
y-5/7=3x
divide both sides by 3
$\frac{y- \frac{5}{7} }{3}$ =x
replace with f⁻¹(x)

f⁻¹(x)= $\frac{y- \frac{5}{7} }{3}$

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3 years ago

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How do you simplify 4^-2h^-4j^3

notka56 [123]

j^3
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4^2 h^4

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3 years ago

Explain how to find each root, with its multiplicity, for the polynomial y=x^5-27x^2.

Nuetrik [128]

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

The characteristics of the polynomial can be derived from algebraic techniques. Roots and multiplicity can be found by <em>factoring</em> the polynomial. The multiplicity is represented by a power binomial of the form:

$(x-r_{i})^{m},\,m \le n$ (1)

Where $n$ is the grade of the polynomial.

Now we proceed to factor the formula:

$y = x^{5}-27\cdot x^{2}$

$y = x^{2}\cdot (x^{3}-27)$

$y = x^{2}\cdot (x^{2}+3\cdot x +9)\cdot (x-3)$

Please notice that $x^{2}+3\cdot x + 9$ have two complex roots.

$y = x^{2}\cdot \left(x+\frac{3}{2}-i\,\frac{3\sqrt{3}}{2} \right)\cdot \left(x+\frac{3}{2}+i\,\frac{3\sqrt{3}}{3} \right)\cdot (x-3)$

The roots of $y = x^{5}-27\cdot x^{2}$ are 0 (multiplicity 2), -3 (multiplicity 1), $-\frac{3}{2}+i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1) and $-\frac{3}{2}-i\,\frac{3\sqrt{3}}{2}$ (multiplicity 1).

We kindly invite to see this question on polynomials: brainly.com/question/1218505

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2 years ago